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Chapter 22
Current and Resistance
Topics:
• Current
• Conservation of current
• Batteries
• Resistance and resistivity
• Simple circuits
Sample question:
How can the measurement of an electric current passed through a
person’s body allow a determination of the percentage body fat?
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Slide 22-1
Series and Parallel
Series
• Circuit elements in a chain between 2 points
• Same current flows through circuit elements
I1 = I2
• Electric Potentials add => Delta Vtotal = Delta V1 + Delta V2
Parallel
• Circuit elements on multiple paths connecting the same
points
• Since paths connect the same points,
Delta V’s are the same
• Currents Add => Itotal = I1 + I2
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Slide 22-25
Show equipotentials in a circuit
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Slide 22-35
Example Problem
The filament of a 100-W bulb carries a current of 0.83 A at the
normal operating voltage of 120 V.
A.
B.
What is the resistance of the filament?
If the filament is made of tungsten wire of diameter 0.035
mm, how long is the filament?
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Slide 22-28
Why is one bulb Brighter, 40 W vs. 100 W
Why is one bulb brighter?
Which has the greatest resistance?
In parallel, which carries the greatest current?
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Slide 22-35
Which bulb is brighter in series, 40 W vs. 100 W
Why is one bulb brighter?
Which has the greatest resistance?
In parallel, which carries the greatest current?
How are the plugs in your home wired?
A. In series
B. In parallel
C. In combination
How can you tell?
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Slide 22-35
Conductivity and Resistivity
This woman is measuring her percentage body fat by
gripping a device that sends a small electric current
through her body. Because muscle and fat have different
resistivities, the amount of current allows the fat-to-muscle
ratio to be determined.
Slide 30-66
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Example Problem: Measuring Body Fat
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Slide 22-35
Analyzing a Simple Circuit
•
•
In a circuit using a battery, a lightbulb, and wires (a flashlight),
the lightbulb has a resistance of ~3 Ω. The wires typically have
a much lower resistance.
We use an ideal wire where its resistance is 0. The potential
difference in the wire is 0, even if there is current in it.
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Analyzing a Simple Circuit
• For the ideal-wire
model, two wires are
connected to a
resistor. Current flows
through all three, but
the current only
requires a potential
difference across the
resistor.
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Analyzing a Simple Circuit
• Current moves in the
direction of
decreasing potential,
so there is a voltage
drop when the
current passes
through the resistor
left to right.
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Analyzing a Simple Circuit
•
The electric field in a resistor carrying a current in a
circuit is uniform. The strength of the electric field is
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Charge Carriers
 The outer electrons of
metal atoms are only
weakly bound to the
nuclei.
 In a metal, the outer
electrons become
detached from their
parent nuclei to form a
fluid-like sea of electrons
that can move through the
solid.
 Electrons are the
charge carriers in
metals.
Slide 30-22
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Discharging a Capacitor
 The wire is already full
of electrons!
 We don’t have to wait
for electrons to move
all the way through the
wire from one plate to
another.
 We just need to slightly
rearrange the charges
on the plates and in
the wire.
Slide 30-30
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Establishing the Electric Field in a Wire
 The figure shows
two metal wires
attached to the
plates of a charged
capacitor.
 This is an
electrostatic
situation.
 What will happen if
we connect the
bottom ends of the
wires together?
Slide 30-32
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Establishing the Electric Field in a Wire
 Within a very brief
interval of time
(<10-9 s) of
connecting the wires,
the sea of electrons
shifts slightly.
 The surface charge is
rearranged into a
nonuniform
distribution, as shown
in the figure.
Slide 30-33
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Establishing the Electric Field in a Wire
 The nonuniform distribution of surface charges along a
wire creates a net electric field inside the wire that
points from the more positive end toward the more
negative end of the wire.
 This is the internal electric field that pushes the electron
current through the wire.
Slide 30-34
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QuickCheck 30.2
Surface charge is distributed on a wire as shown.
Electrons in the wire
A. Drift to the right.
B. Drift to the left.
C. Move upward.
D. Move downward.
E. On average, remain at
rest.
Slide 30-35
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QuickCheck 30.2
Surface charge is distributed on a wire as shown.
Electrons in the wire
A. Drift to the right.
B. Drift to the left.
C. Move upward.
D. Move downward.
Electric field from
nonuniform surface
charges is to the right.
Force on negative
electrons is to the left.
E. On average, remain at
rest.
Slide 30-36
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A Model of Conduction
 Within a conductor
in electrostatic
equilibrium, there is
no electric field.
 In this case, an
electron bounces
back and forth
between collisions,
but its average
velocity is zero.
Slide 30-37
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A Model of Conduction
 In the presence of an
electric field, the
electric force causes
electrons to move
along parabolic
trajectories between
collisions.
 Because of the
curvature of the
trajectories, there is a
slow net motion in the
“downhill” direction.
Slide 30-38
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A Model of Conduction
 The graph
shows the
speed of an
electron during
multiple
collisions.
 The average
velocity is the
electron drift
speed
Slide 30-39
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Chapter 23
Circuits
Topics:
• Circuits containing multiple
•
•
•
elements
Series and parallel
combinations
RC circuits
Electricity in the nervous
system
Sample question:
An electric eel can develop a potential difference of over 600 V. How
do the cells of the electric eel’s body generate such a large potential
difference?
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Slide 23-1
Kirchhoff’s Laws
• Kirchhoff’s junction law, as we
learned in Chapter 22, states that
the total current into a junction
must equal the total current
leaving the junction.
• This is a result of charge and
current conservation:
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Kirchhoff’s Laws
•
•
•
The gravitational potential energy of an object
depends only on its position, not on the path it
took to get to that position.
The same is true of electric energy. If a
charged particle moves around a closed loop
and returns to its starting point, there is no net
change in its electric energy: Δuelec = 0.
Because V = Uelec/q, the net change in the
electric potential around any loop or closed
path must be zero as well.
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Kirchhoff’s Laws
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Kirchhoff’s Laws
•
•
•
For any circuit, if we add all
of the potential differences
around the loop formed by
the circuit, the sum must
be zero.
This result is Kirchhoff’s
loop law:
ΔVi is the potential difference of the ith component of the loop.
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Kirchhoff’s Laws
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Slide 23-11
Kirchhoff’s Laws
Text: pp.
729–730
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Kirchhoff’s Laws
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Text: p.
730
Kirchhoff’s Laws
• ΔV
bat can
be positive or negative for a
battery, but ΔVR for a resistor is always
negative because the potential in a
resistor decreases along the direction of
the current.
• Because the potential across a resistor
always decreases, we often speak of the
voltage drop across the resistor.
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Circuit Analysis using Kirchoff’s rules (Loop +
Junction)
Circuit from End of Circuit Activity 1
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Slide 22-35
QuickCheck 23.6
The diagram below shows a segment of a circuit. What is
the current in the 200 Ω resistor?
•
0.5 A
•
1.0 A
•
1.5 A
•
2.0 A
•
There is not enough
information to decide.
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QuickCheck 23.6
• The diagram below shows a segment of a
circuit. What is the current in the 200 Ω
resistor?
• 0.5 A
• 1.0 A
• 1.5 A
• 2.0 A
• There is not enough information to
decide.
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Series and Parallel
• There are two
possible ways that you
Circuits
can connect the circuit.
• Series and parallel circuits have very
different properties.
• We say two bulbs are connected in
series if they are connected directly to
each other with no junction in between.
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Calculating Equivalent Resistance
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Slide 22-35
Parallel Resistors
• The current I
from the battery splits into
currents I1 and I2 at the top of the
junction.
bat
• According to the junction law,
• Applying Ohm’s law to each resistor, we
find that the battery current is
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•
•
•
Parallel Resistors
Can we replace a group of parallel resistors with a
single equivalent resistor?
To be equivalent, ΔV must equal ℇ and I must equal
Ibat:
This is the equivalent resistance, so a single Req acts
exactly the same as multiple resistors.
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Parallel Resistors
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Equivalent Resistance Examples
1. By using only two resistors singly, in series, or in parallel you are
able to obtain resistances of 18.0, 24.0, 72, and 96 . What are
the two resistances? (Make sure your answer is consistent.)
2. Find the equivalent resistance of the following circuit: Be sure to
show each step.
3. Find the heat energy generated by the 3 Ohm and the 12 Ohm
resistors in the circuit in problem 2
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Slide 22-35